the fiber product $\underline{X}\times_{\mathcal{D}}\underline{M}$ is coming from a manifold i.e., it is isomorphic to stack of the form $\underline{P}$ for some smooth manifold $P$ and the map $P\rightarrow X$ coming from the projection map $\underline{X}\times_{\mathcal{D}}\underline{M}\rightarrow \underline{M}$ is a smooth submersion.

I understand the set up. What I do not understand is how do you check if something is an atlas for a stack or more simpler is, how do you check if a stack $\mathcal{C}$ is isomorphic to $\underline{P}$ for some manifold.

1 Answer
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Take $M = X$ and $f=p$, so that $\underline{P} = \underline{X} \times_\mathcal{D} \underline{X}$. The Lie groupoid $P\rightrightarrows X$ you get should be proper, in the sense the source-target map $(s,t)\colon P\to X\times X$ is a proper map, and $(s,t)$ should be injective, so that there are no nontrivial automorphism groups. Then your original stack is the stack associated to the manifold that is the quotient of $X$ by the equivalence relation corresponding to $P$ (which is closed, by properness of $(s,t)$), i.e. $\mathcal{D} \simeq \underline{X/P}$.

To respond to the question about non-injectivity of $(s,t)$ in the comments:

Let us say that the groupoid $P \rightrightarrows X$ has some pair $x,y\in X$ are such that $(s,t)^{-1}(x,y)$ has at least two distinct elements, say $a$ and $b$. Then note that $id_x$ and $b^{-1}a$ are distinct elements in $(s,t)^{-1}(x,x)=\mathrm{Aut}(x)$. Then consider the groupoid of functors $\ast \to (P\rightrightarrows X)$ and natural isomorphisms, which is equivalent to the groupoid of functors $\ast \to \mathcal{D}$ (this is because every principal $(P \rightrightarrows X)$-bundle over $\ast$ has a section - this is a nice exercise to work out the details). But by Yoneda the groupoid of functors $\ast \to \mathcal{D}$ is equivalent to the groupoid $\mathcal{D}(\ast)$, thinking of $\mathcal{D}$ as a functor $\mathbf{Mfld}^\mathrm{op}\to \mathbf{Gpd}$. But in the groupoid of functors $\ast\to (P\rightrightarrows X)$ the functor sending $\ast$ to $x\in X$ has a nontrivial automorphism, given by $b^{-1}a$, so the groupoid $\mathcal{D}(\ast)$ has an object with a nontrivial automorphism, hence $\mathcal{D}\colon \mathbf{Mfld}^\mathrm{op}\to \mathbf{Gpd}$ doesn't factor through $\mathbf{Set}\hookrightarrow \mathbf{Gpd}$. Every stack arising from a manifold is in fact a sheaf $\mathbf{Mfld}^\mathrm{op}\to \mathbf{Set}$, so in this case $\mathcal{D}$ cannot be the stack associated to a manifold. Hence for $\mathcal{D}$ to come from a manifold, $(s,t)$ must be injective. This is not sufficient, one really does need properness else the quotient $X/P$, which should be the manifold in question, will not exist in $\mathbf{Mfld}$. Consider for instance the action groupoid associated to a free action of $\mathbb{R}$ on a torus $U(1)^2$ with irrational slope.

$\begingroup$Thanks for your answer. I am trying to understand this.. Question is, given a geometric stack $\mathcal{D}$, how do you know if that stack comes from a manifold.. What you are saying is, take an atlas for $\mathcal{D}$ say $X\rightarrow \mathcal{D}$ and consider fiber product with itself, giving a stack, which we know is coming from a manifold, you are naming it $P$. You get a Lie groupoid $P\rightrightarrows X$..$\endgroup$
– Praphulla KoushikAug 15 '18 at 10:01

$\begingroup$If the map $(s,t)$ is proper and injective, then you are saying $\mathcal{D}$ comes from manifold $X/P$ where the relation is $x\sim y$ if there exists $p\in P$ such that $(s,t)(p)=(x,y)$.. Am I correct?? How did you think of this? Can you give some reference.. please..$\endgroup$
– Praphulla KoushikAug 15 '18 at 10:01

$\begingroup$The geometric stack is equivalent to a manifold if the Lie groupoid I gave is equivalent to a manifold in the sense in my 2012 paper. (This is because the 2-categories of geometric stacks and internal groupoids with anafunctors as maps are equivalent.) That is the case precisely when (s,t) is injective and the quotient of the resulting equivalence relation is a manifold. This is the case precisely when (s,t) is closed (then properness comes for free), a result going back to Godement, I believe (see eg math.stackexchange.com/q/496571/3835 )$\endgroup$
– David RobertsAug 15 '18 at 10:17

$\begingroup$Also, if (s,t) is not injective you know instantly that you don't have a manifold as then you have genuinely stacky points.$\endgroup$
– David RobertsAug 15 '18 at 10:28

$\begingroup$Oh, ok ok.. Can you tell me title of your 2012 paper.. there seem to be two...$\endgroup$
– Praphulla KoushikAug 15 '18 at 12:02